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[tex]\boxed{(a+b)^n=\sum\limits_{p=0}^{n}\binom{n}{p}a^{n-p}b^p}[/tex]
[tex]\boxed{T_{p+1}=\binom{n}{p}a^{n-p}b^p}[/tex]
O binômio é:
[tex]\bigg(x+\dfrac{a}{x}\bigg)^6[/tex]
Seu termo geral será:
[tex]T_{p+1}=\dbinom{6}{p}x^{6-p}\bigg(\dfrac{a}{x}\bigg)^p\ \therefore\ \boxed{T_{p+1}=\dbinom{6}{p}x^{2(3-p)}a^p}[/tex]
Para o termo [tex]x^4[/tex], o valor de [tex]p[/tex] será:
[tex]2(3-p)=4\ \therefore\ 3-p=2\ \therefore\ \boxed{p=1}[/tex]
Dessa forma:
[tex]T_{1+1}=\dbinom{6}{1}x^{2(3-1)}a^1\ \therefore\ T_2=\dfrac{6!}{1!5!}x^4a\ \therefore\ T_2=6ax^4[/tex]
Mas [tex]T_2=12x^4[/tex], logo:
[tex]12x^4=6ax^4\ \therefore\ 12=6a\ \therefore\ \boxed{a=2}[/tex]
Resposta: o valor de a é 2.